The slope of line joining \[P\left( {6,k} \right)\] and \[Q\left( {1 - 3k,3} \right)\] is \[\dfrac{1}{2}\] . Find \[\left| k \right|\].

Hint:
We are given two points of a line with their coordinates. Also the slope of that line joining these two given points is \[\dfrac{1}{2}\]. Then to find k use formula to find slope of a line that is given by \[\dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}\] .

Complete step by step solution:
Given two points
 \[P\left( {6,k} \right)\] and \[Q\left( {1 - 3k,3} \right)\]
Let \[P\left( {6,k} \right) = P\left( {{x_1},{y_1}} \right)\]
 And \[Q\left( {1 - 3k,3} \right) = Q\left( {{x_2},{y_2}} \right)\]
Formula to find slope of line joined by these two points is given by
Slope = \[\dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}\]
But the slope of that line is already given and it is \[\dfrac{1}{2}\].
\[
\Rightarrow \dfrac{1}{2} = \dfrac{{3 - k}}{{1 - 3k - 6}} \\
\Rightarrow \dfrac{1}{2} = \dfrac{{3 - k}}{{ - 5 - 3k}} \\
\Rightarrow 1\left( { - 5 - 3k} \right) = 2\left( {3 - k} \right) \\
\Rightarrow - 5 - 3k = 6 - 2k \\
\Rightarrow - 5 - 6 = 3k - 2k \\
\Rightarrow k = - 11 \\
\]
But we need to find the value of modulus of k.
\[\left| k \right| = 11\]

Additional information:
1) Equation of a straight line is given by \[y = mx + c\].
2) Slope of a line is given by letter m generally.
3) It is expressed in the form of a ratio of vertical change to horizontal change.
4) A line is increasing if the slope is positive.
5) A line is decreasing if the slope is negative.
6) If the line is horizontal then the slope of the line is undefined.
7) If the line is vertical then the slope is zero.
8) Slope of line is also given by \[\tan \theta \], where \[\theta \] is the angle of inclination of the line.

Note:
The modulus of k gives k when it is positive and if it is negative then it’ll open with a negative sign which eventually makes k positive. In a nutshell mod never gives negative value.